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Question

PCl5(g)PCl3(g)+Cl2(g)
α is the degree of dissociation of PCl5 at equilibrium pressure P. Which among the following is the correct expression for degree of dissociation of α ?

A
α=KPP+KP
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B
α=P+KPKP
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C
α=KPP
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D
α=P/KP
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Solution

The correct option is A α=KPP+KP
PCl5PCl3+Cl2

1 0 0 (Conc. at starting)

1α α α (At Eqilibrium)

Then total no. of moles =(1α)+α+α=1+α

Let total pressure be P.

Partial pressure of:-

PCl5=[(1α)/(1+α)]×P

PCl3=[(α)/(1+α)]×P

Cl2=[(α)/(1+α)]×P

K(P)=[(α)/(1+α)]×P×[(α)(1+α)]×P]/[(1α)/(1+α)]×P

=α×α×P/(1α2)

α=KPP+KPA is correct answer



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